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Question

Consider the equilibrium of any charge $- \mathrm{Q}.$

$\frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{\sqrt {2{Q^2}} }}{{{a^2}}} + \frac{{{Q^2}}}

Vedclass · Accepted Answer

$(Q/ 4)(1+ 2\sqrt 2)$

Vedclass · Answer

Consider the equilibrium of any charge $- \mathrm{Q}.$

$\frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{\sqrt {2{Q^2}} }}{{{a^2}}} + \frac{{{Q^2}}}{{2{a^2}}}} ight] = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_0}}}{{{{(a/\sqrt 2 )}^2}}}$

$	herefore $ $\frac{Q}{4}=\frac{Q}{4}(1+2 \sqrt{2}).$

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is

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